I often guess generalizations based on subtle patterns. I found the following formula in such a way and I wonder how it can be proven.
Let $n$ be a non-negative integer. Then
\begin{align} \text{Cl}_{n}(x)&=(-1)^{\lfloor\frac{n-1}{2}\rfloor}x^{n-1}\left(\frac{H_{n-1}-\ln x}{\Gamma(n)}+\sum_{k=1}^{\lfloor\frac{n-1}{2}\rfloor}(-1)^k\frac{\zeta(2k+1)}{x^{2k}\Gamma(n-2k)}+2\sum_{k=1}^\infty\frac{\zeta(2k)\Gamma(2k)}{\Gamma(n+2k)}\left(\frac{x}{2\pi}\right)^{2k}\right) \\&=(-1)^{\lfloor\frac{n-1}{2}\rfloor}\left(\psi^{(-n)}(x)-\psi^{(-n)}(x+1)+\sum_{k=1}^{\lfloor\frac{n-1}{2}\rfloor}(-1)^k\frac{x^{n-2k-1}\zeta(2k+1)}{\Gamma(n-2k)}+2\sum_{k=1}^\infty\frac{x^{n+2k-1}\zeta(2k)\Gamma(2k)}{(2\pi)^{2k}\Gamma(n+2k)}\right) \end{align}
where $\text{Cl}_{n}(x)$ is one of the standard Clausen functions defined by
$$ \text{Cl}_{n}(x)= \begin{cases} \sum_{k=1}^\infty\frac{\sin(kx)}{k^n}& n\space\text{even}\\\\ \sum_{k=1}^\infty\frac{\cos(kx)}{k^n}& n\space\text{odd} \end{cases} $$
Edit: Upon further investigation, the proof can likely be derived from equations 2.7 and 2.8 of this paper.